Question: Is ${292396}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {292396}= &&{2}\cdot100000+ \\&&{9}\cdot10000+ \\&&{2}\cdot1000+ \\&&{3}\cdot100+ \\&&{9}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {292396}= &&{2}(99999+1)+ \\&&{9}(9999+1)+ \\&&{2}(999+1)+ \\&&{3}(99+1)+ \\&&{9}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {292396}= &&\gray{2\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {2}+{9}+{2}+{3}+{9}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${292396}$ is divisible by $3$ if ${ 2}+{9}+{2}+{3}+{9}+{6}$ is divisible by $3$ Add the digits of ${292396}$ $ {2}+{9}+{2}+{3}+{9}+{6} = {31} $ If ${31}$ is divisible by $3$ , then ${292396}$ must also be divisible by $3$ ${31}$ is not divisible by $3$, therefore ${292396}$ must not be divisible by $3$.